Theorem: Assume that $G$ is a group and $H_1,\dots,H_n$ are subgroups of $G$. Define $H:=H_1\dots H_n$ (The direct product) . The statements written below are equivalent:
(i) $H_1\times \dots\times H_n$ is isomorphic to $H$ under the isomorphism $(h_1,\dots,h_n)\mapsto h_1\dots h_n$
(ii) $H_i$ is a normal subgroup of $H$ and each member of $H$ has a unique representation like $h_1\dots h_n$ such that $\forall i \space h_i\in H_i$
(iii) $H_i$ is a normal subgroup of $H$ and if $e$ has a unique representation like $e=x_1\dots x_n$ , then $x_1=\dots=x_n=e$
(iv) $H_i$ is a normal subgroup of $H$ and for all $i = 1,\dots,n$, We have: $H_i \cap (H_1\dots \hat H_i\dots H_n)=\{e\}$ . (The hat sign means that $H_i$ is omitted in the product... something like $H_1\dots H_{i-1} H_{i+1}\dots H_n$)
We have proved the theorem above in the class. I was thinking if we can consider another statement which is equivalent to those four statements. My statement is :
(v) $H_i$ is a normal subgroup of $H$ and $\forall i \in \{1,\dots,n\}\quad |H_1\dots H_n|=|H_1|\dots|H_n|$
I think that we can reach (v) from (i). But, Can we reach (i) from (v), too?
Any other way to show that those four are equivalent to (v) will be appreciated. Even if u think that this is not correct, provide a counterexample.
Yes, (v) is also equivalent to (i) (assuming that each $H_i$ is finite). But it's easier to simply prove that (v) implies (ii). To see it, consider the function $$\varphi:H_1\times \cdots \times H_n\to H : (h_1,\dots,h_n)\mapsto h_1\dots h_n.$$ By definition of the internal product, it is surjective. By the second part of (v), its domain and codomain are finite sets with the same cardinality, so it is also injective, which is exactly what (ii) says.